Inequality condition for real numbers

We are given: $\frac{x+3}{y-3} - \frac{x}{y} > 0$ $\frac{y(x+3) - x(y-3)}{y(y-3)} > 0$ $\frac{xy + 3y - xy + 3x}{y(y-3)} > 0 \implies \frac{3(x+y)}{y(y-3)} > 0$

So $\frac{x+y}{y(y-3)} > 0$.

If $x < 0$:

• If $y > 3$, then $y(y-3) > 0 \implies x + y > 0 \implies y > -x > 0$. In this case, $-x < y$.
• If $0 < y < 3$, then $y(y-3) < 0 \implies x + y < 0 \implies -x > y$.
• If $y < 0$, then $y(y-3) > 0 \implies x + y > 0 \implies x > -y$. But since $x < 0$ and $y < 0$, $-y > 0$, so $x > -y$ is impossible.

Thus, if $x < 0$, $y$ must lie in $(0, 3)$, which forces $-x > y$.

Hence, Option B ("if $x < 0$, then $-x > y$") is correct.